Electromagnetic Waves Important Questions.

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1. Q1Medium

   A prism has an angle of minimum deviation of 40° and a refracting angle of 60°. What is the refractive index of the prism material?

   • A.1.33
   • B.1.53
   • C.1.73
   • D.2.00
   Answer: 1.53

   Explanation: Using the formula for minimum deviation in a prism: n = sin[(A + δm)/2] / sin(A/2), where A is the prism angle and δm is the angle of minimum deviation. Here, n = sin[(60° + 40°)/2] / sin(60°/2) ≈ 1.53.

2. Q2Medium

   If the prism from the previous question is placed in water (refractive index 1.33), what will be the new angle of minimum deviation?

   • A.20°
   • B.30°
   • C.40°
   • D.50°
   Answer: 30°

   Explanation: The angle of minimum deviation changes when the prism is immersed in water due to the change in relative refractive index. The formula for minimum deviation in a new medium is adjusted accordingly, resulting in an approximate angle of 30°.

3. Q3Medium

   For a double-convex lens made from glass with a refractive index of 1.55, and both faces having the same radius of curvature, what radius of curvature is required if the focal length is to be 20 cm?

   • A.20 cm
   • B.25 cm
   • C.30 cm
   • D.35 cm
   Answer: 30 cm

   Explanation: Using the lensmaker’s formula, 1/f = (n-1)(1/R1 - 1/R2). For a double-convex lens with equal radii, R1 = R2 = R. Solving for R, we get R = 2f(n-1) = 2 * 20 cm * (1.55 - 1) = 31 cm (approx).

4. Q4Medium

   A beam of light converges at a point P. A convex lens of focal length 20 cm is placed 12 cm from P. Where will the beam converge after passing through the lens?

   • A.12 cm on the same side of the lens
   • B.20 cm on the same side of the lens
   • C.12 cm on the other side of the lens
   • D.20 cm on the other side of the lens
   Answer: 12 cm on the other side of the lens

   Explanation: Since the beam converges at point P, which is 12 cm from the lens, and the focal length of the lens is 20 cm, the lens will bring the converging rays to a focus 12 cm on the other side of the lens.

5. Q5Medium

   A beam of light converges at a point P. A concave lens of focal length 16 cm is placed 12 cm from P. Where will the beam converge after passing through the lens?

   • A.The beam will converge at 16 cm on the same side
   • B.The beam will converge at 24 cm on the same side
   • C.The beam will diverge as if coming from 24 cm on the same side
   • D.The beam will diverge as if coming from 12 cm on the opposite side
   Answer: The beam will diverge as if coming from 24 cm on the same side

   Explanation: Using the lens formula 1/v - 1/u = 1/f, where u = -12 cm and f = -16 cm, we get v = -24 cm, indicating the beam diverges as if originating from 24 cm on the same side.

6. Q6Medium

   What is the critical angle for total internal reflection when light travels from glass (refractive index 1.5) to water (refractive index 1.33)?

   • A.48.79°
   • B.61.04°
   • C.70.53°
   • D.82.82°
   Answer: 61.04°

   Explanation: Using the formula for critical angle, sin(θ_c) = n2/n1, where n1 = 1.5 and n2 = 1.33. Solving for θ_c gives sin⁻¹(1.33/1.5) ≈ 61.04°.

7. Q7Medium

   A convex mirror has a focal length of 15 cm. If an object is placed 12 cm away from the mirror, where will the image be formed?

   • A.6 cm in front of the mirror
   • B.12 cm behind the mirror
   • C.6 cm behind the mirror
   • D.15 cm behind the mirror
   Answer: 6 cm behind the mirror

   Explanation: Using the mirror equation, 1/f = 1/v + 1/u, where f = 15 cm and u = -12 cm. Solving for v gives v ≈ 6 cm (positive value indicates a virtual image behind the mirror).

8. Q8Medium

   What is the magnification produced by the convex mirror in the previous question?

   • A.-0.5
   • B.0.5
   • C.1
   • D.2
   Answer: -0.5

   Explanation: Magnification, m = -v/u = -6 cm / -12 cm = 0.5 (negative sign indicates the image is virtual and erect).

9. Q9Medium

   A light ray travels from air (refractive index 1) into a glass slab (refractive index 1.5). If the angle of incidence is 30°, what is the angle of refraction?

   • A.10°
   • B.19.47°
   • C.30°
   • D.45°
   Answer: 19.47°

   Explanation: Using Snell's Law, n1*sin(θ1) = n2*sin(θ2). Solving for θ2 gives sin⁻¹((1*sin(30°))/1.5) ≈ 19.47°.

10. Q10Medium

    What is the power of a lens with a focal length of 0.5 meters?

    • A.1 diopter
    • B.2 diopters
    • C.0.5 diopters
    • D.0.25 diopters
    Answer: 2 diopters

    Explanation: Using the formula for lens power, P = 1/f, where f is in meters. P = 1/0.5 = 2 diopters.

11. Q11Medium

    A needle is placed at the bottom of a tank filled with water to a height of 12.5 cm. The apparent depth of the needle is measured to be 9.4 cm. What is the refractive index of water?

    • A.1.25
    • B.1.33
    • C.1.5
    • D.1.75
    Answer: 1.33

    Explanation: Using the formula for apparent depth, n = real depth / apparent depth. Therefore, n = 12.5 cm / 9.4 cm ≈ 1.33.

12. Q12Medium

    A telescope has an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope?

    • A.10
    • B.20
    • C.25
    • D.50
    Answer: 20

    Explanation: The magnifying power of a telescope is given by the ratio of the focal lengths of the objective and eyepiece. Therefore, magnifying power = 100 cm / 5 cm = 20.

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